How can I create hindrances to radio waves?
[Physics] How to create hindrances to radio waves
electromagnetic-radiation
Related Solutions
Consider the incoming electric field of the radio waves. This field is a superposition of all broadcasts from stations near your receiver. The job of the receiver is to pick out one of these transmissions and turn it into sound.
AM radio
Now consider an AM radio station transmitter. Suppose the sound wave that station wants to transmit is represented by a function of time $m(t)$ where here $m$ is for "message". Note that $m(t)$ includes all information about the sound, i.e. it includes frequency, amplitude... everything. In an AM transmitter, we use a circuit to multiply $m(t)$ by a sinusoid, creating the transmitted signal $$s(t) = m(t) \cos(\Omega t)$$ where here $s$ stands for "signal" and $\Omega$ is called the "carrier frequency". Here we see the reason for the term Amplitude Modulation (AM): the message is a modulation of the amplitude of the carrier wave.
You can use trig identities or Fourier analysis to see that the spectral content of $s(t)$ is in the range $\Omega \pm \delta \omega$ where $\delta \omega$ is the highest frequency in $m(t)$. The carrier frequency $\Omega$ might be in the tens of MHz range. On the other hand, the actual message $m(t)$ would absolutely never have any frequencies above around 20 kHz because that's the upper range of human hearing. In real life, $m(t)$ doesn't use up the full 20 kHz; useful speech and music don't need our full hearing range.
So now we see that the transmitted signal $s(t)$ is contained within some relatively narrow bandwidth, i.e. maybe a 10 kHz band centered at 10 MHz. Therefore, a tuned circuit with a $Q$ of around 1,000 and centered at $\Omega$ picks up $s(t)$ but mostly nothing else.$^{[a]}$ Of course, we also have to enforce that the various stations' carrier frequencies are separated by more than their $\delta \omega$'s so that nobody's transmissions overlap with anyone else's.
So, the output of our tuned circuit is roughly just $s(t)$! I say "roughly" because our tuned circuit isn't perfect, so we might pick up a bit of stuff from other transmissions, but since it's farther away from the center of our tuned circuit the amplitude is suppressed. Then, we just put the signal through a rectifier and a low pass filter so that the carrier oscillations are gone and we only get $m(t)$. That's it! Now we have the original sound message and we can put it into a speaker. We don't have to think about amplitude and frequency separately: we have the entire original sound waveform.
$[a]$: $Q$ is the center frequency divided by the bandwidth, so $$Q = 10 \text{MHz} / 10 \text{kHz} = 1,000 \, .$$
They are self propagating. An oscillation in the electric field results in an induced one in the magnetic field, and vice versa repeatedly. Thus, it propagates itself through space with these repeated inductions regardless of physical surroundings.
See http://www.physics.usyd.edu.au/teach_res/hsp/u7/t7_emr.htm.
Best Answer
Any conductive material will block any EM radiation, including radio waves. However, the thickness of material required to fully block the radiation depends on the wavelength of the radiation and the properties of the material. With a perfect conductor, the radiation is stopped completely by any thickness of the material, but in any real material (except for a superconductor) the radiation is attenuated as it travels through the material. The strength of the radiation will fall off with an exponential decay, and it will do so more quickly in a better conductor. The important number here is called "skin depth" which is given by:
$$\delta = \frac{c}{\sqrt{2\pi\sigma\mu\omega}} $$
Where $\sigma$ is the conductivity of the material, $\mu$ is the magnetic permeability of the material, and $\omega$ is the frequency of the radiation. The skin depth is the distance into the material at which the radiation will have fallen to a fraction of $1/e \approx 0.37$ relative to its original intensity.
As you can see, the skin depth will be smaller for a better conductor or higher frequency radiation. For example, for Aluminum, radiation at 60Hz will have a skin depth of over 8 millimeters, while optical radiation around 600 THz will have a skin depth around 3 nanometers! This is why metal blocks visible light so well, yet radio waves easily penetrate into your house.
Sources:
"Lectures on Electromagnetism" page 280. Ashok Das; Hindustan Book Agency, New Delhi, 2004.
http://en.wikipedia.org/wiki/Skin_effect